OpAmp Voter

Three opamps, each with a diode in its output. Each acts like a follower, but all three get their nfb from the output node. Needs a pulldown from output to ground or V-, too.

Classic process control "high select."

You're welcome.

Cheers

nd pass the highest of the signals, without diode losses. Using a micropro cessor is out of the question. Cost is a object and actives should be limi ted to devices that can be second sourced.

Yup. A less precise option is putting each input thru a diode, with the out put of the diodes going thru an opamp with a diode in the nfb path to add a diode drop to the output. An extra resistor to supply rail needed to keep the diode biased. 4 diodes 1 opamp versus 3 diodes 3 opamps. The performanc e difference between this and the above should be fairly obvious.

NT

Just add another input to this circuit I posted last November...

bitrexCircuitChallenge_2013_11_07.pdf

on the S.E.D/Schematics Page of my website.

Probably best if you used a quad-OpAmp... three for inputs, the resistor to ground (or to negative rail) to ensure "Who's on first"

Fourth OpAmp as the buffer. ...Jim Thompson

With something like this tacked on it will work better at 5kHz:

It will still be a bit glitchy where the levels cross.

Best regards, Spehro Pefhany

Two of the three amps will be railed negative at any time, so your circuit doesn't really need R2.

I'd go for schottky diodes and faster amps, meself.

At 5kHz, any of the suggestions above will do.

I also have one that works at high frequencies, without suffering from glitches or integrator wind-up.

Tim

-- Seven Transistor Labs Electrical Engineering Consultation Website:

This is for converting a RGB command signal to an intensity signal.

Steve

The reverse diodes + 10K resistors keep the op-amps from railing. Some op-amps take a long time to recover from being railed (as I think you have measured).

Using Schottky diodes as you suggest would reduce the volts they have to slew.

Best regards, Spehro Pefhany

No, No, don't show us that one.

"I don't expect you to look, Mr. Bond, I expect you to die."

Tim

Old-school:

V+ -+- | .--+--. | | [R1] | | |/ .---+---| Q2 | |>. | Vout |\ | | . --- . V-

Don't forget the output load...

Cheers, James

Works okay provided BE junction of Q2 doesn't break down, should be fine for 5 or 6V with just about any transistor.

I guess you could flip Q2 for more voltage- might even reduce the offset.

Best regards, Spehro Pefhany

Q2 shouldn't break down--video's only +/- 1v.

You mean use Q2 inverted? Wouldn't r(e) have the same i*r drop?

Using a compound (Szilaki?) for Q2 would reduce the gain error, but it gets messy. Using the usual feedback resistors you could tweak Q5's gain slightly over unity. (Hey, it's only video, right?)

V+ -+- | .------+------. | | | | [R2] | | | |. | v1 >------| | | | |\ | +------+----> Q1 | | | --- | [RL] V- | | .--+ --- |------------| |\ Q3 | --- V-

Op-amps are almost always the way to go, but I always enjoy remembering the simpler things...

Cheers, James Arthur

Ooops, I inverted that into a lowest-wins voter! Flip all the polarities, swap npnpnp, etc., please.

Cheers, James Arthur

Best regards,

--sp

You can also use transistors instead of diodes, and wire-OR their emitters. That usually makes the op amps better behaved IME. A catch diode (as Spehro shows) helps with the transient.

Cheers

Phil Hobbs

On a sunny day (Mon, 8 Sep 2014 13:16:11 -0700 (PDT)) it happened " snipped-for-privacy@SpamSucks.net" wrote in :

In the European system the ratio for white is .3 R + .59 G + .11 B

You can nicely appoximate this like this:

R -- 3k9 -- |-------- BW G -- 1k8 --| | B -- 10k --

Note that for R, B, G = 1V the output is also 1V.

The rest is, as they say, bullshit.

Your question sucks, so this is a wild guess of what you were trying to accomplish, or not accomplish.

So be more clear in your questions! That saves people time and irritation. Or google first.

El Pante

If you need super-fast, and no OpAmp-style hangs, study the techniques in Patent 3,643,110 Sample and Hold Circuit, PDF on the Home Page of my website.

All you need is the first half of the schematic.

How it works is detailed in the Patent. ...Jim Thompson

Use the right transistor:

this one takes 25V across the BE junction.